Python线程间的同步与互斥

来源:互联网 发布:宁波奥派网络 编辑:程序博客网 时间:2024/06/01 22:19

最简单来说,假如有3个线程同时访问一个全局变量,那么很可能会操作互斥错误的情况,代码如下:

#!/usr/bin/python#encoding=utf-8import timeimport threadingindex = 0def printIndex():    global index    global lock    while index < 100:        print index,        index += 1        time.sleep(1)t1 = threading.Thread(target = printIndex, args=())t1.setDaemon(True)t1.start()t2 = threading.Thread(target = printIndex, args=())t2.setDaemon(True)t2.start()t3 = threading.Thread(target = printIndex, args=())t3.setDaemon(True)t3.start()time.sleep(10)

在没有互斥的情况下3个线程同时访问一个全局变量,最后结果如下:


由此可见,输出的完全是乱码,因为这个地方除了全局变量是临界资源外控制台也是一个临界资源。所以需要互斥操作。将程序改为:

#!/usr/bin/python#encoding=utf-8import timeimport threadingindex = 0lock = threading.RLock()def printIndex():    global index    global lock    while index < 100:        lock.acquire()        print(index),        index += 1        lock.release()        time.sleep(1)t1 = threading.Thread(target = printIndex, args=())t1.setDaemon(True)t1.start()t2 = threading.Thread(target = printIndex, args=())t2.setDaemon(True)t2.start()t3 = threading.Thread(target = printIndex, args=())t3.setDaemon(True)t3.start()time.sleep(100)
最后输出结果:


由此可见,输出时完全按顺序的。在代码中先用RLock()创建一个锁,当需要访问临界资源时进行加锁lock.require(),等退出时再释放锁lock.release()。

对于临界资源来说,用锁可以很方便实现互斥,而如果要实现同步,则需要用event。


假如程序中需要进行复杂的计算,现在将这计算在另一线程中运行,但是主程序怎么才能知道计算完成了呢?实例代码如下:

import timeimport threadingdef getSum(n, sum):    for i in range(0, n + 1):        sum[0] += i        time.sleep(0.1)     def main():    sum = [0]    t = threading.Thread(target = getSum, args=(100, sum,))    t.setDaemon(True)    t.start()    time.sleep(2)    print('sum = ', sum[0])    if __name__ == '__main__':    main()
现在需要求1 + 2 + 3 + ... + N 的和,计算放在单独线程中进行,但是我们不知道计算什么时候结束,只能假设2s,最后结果如下:


最后结果sum = 153,这显然是不正确的,即使是正确,也是巧合。这时就需要一个类似于信号的东西,让子进程结束后能够通过这个信号告诉主进程,event就可以配上用

场了。event和lock的用法差不多,下面是代码:

import timeimport threadingdef getSum(n, sum, event):    for i in range(0, n + 1):        sum[0] += i        time.sleep(0.1)    event .set()def main():    sum = [0]    event = threading.Event()    t = threading.Thread(target = getSum, args=(100, sum, event))    t.setDaemon(True)    t.start()    event.wait()    print('sum = ', sum[0])    if __name__ == '__main__':    main()
程序中用了event作为一个信号量,初始是没信号的,当子进程完成计算后将其设置为有信号,主进程wait返回,最后结果:



最后,写个生产者和消费者的模型吧:


import timeimport randomimport threadingdef producter(buffer, event):    while True:        event.wait()        event.clear()        if len(buffer) > 4:            print("the buffer is full! producter wait...")        else:            data = random.randint(1, 10)            print('the producter has put data: ', data)            buffer.append(data)        event.set()        time.sleep(0.8)    pass    def consumer(buffer, event):    while True:        event.wait()        event.clear()        if len(buffer) == 0:            print('he buffer is empty! consumer wait...')        else:            data = buffer[0]            del buffer[0]            print('the consumer has get data:', data)        event.set()        time.sleep(0.4)def main():    buffer = []    event = threading.Event()    event.set()    p = threading.Thread(target = producter, args=(buffer, event))    p.setDaemon(True)    p.start()    c = threading.Thread(target = consumer, args=(buffer, event))    c.setDaemon(True)    c.start()    if __name__ == '__main__':    main()    time.sleep(100)

这儿有个问题,这里是wait之后再手工clear的,那么这中间就有可能出现线程切换,有没有方法在wait返回后自动clear的呢??


0 0
原创粉丝点击